Development of complexes with a Multiple Bond between Transition Metal and Phosphor us

过渡金属和磷之间具有多重键的配合物的开发

基本信息

批准号：
10440195
负责人：
MIYOSHI Katsuhiko
金额：
$ 8.58万
依托单位：
Hiroshima University
依托单位国家：
日本
项目类别：
Grant-in-Aid for Scientific Research (B)
财政年份：
1998
资助国家：
日本
起止时间：
1998 至 1999
项目状态：
已结题

项目摘要

The reaction of [Cp(CO)ィイD22ィエD2(SnィイD1nィエD1BuィイD23ィエD2)W{PN(Me)CHィイD22ィエD2CHィイD22ィエD2NMe(OMe)}] with a Lewis acid revealed the formation of a phosphenium complex, followed by migration of a ィイD1nィエD1BU group from Sn to P to give a stannylene complex. If the reaction takes place in the presence of NaBPhィイD24ィエD2, a phosphi de complex, Cp(CO)ィイD22ィエD2W{PN(Me)CHィイD22ィエD2CHィイD22ィエD2NMe} is formed. This unprecedented reaction is considered to be a W-P bond order increment by one with elimination of a stannyl and a methoxy group. The reaction proceeds with one electron reducti on of a phosphenium complex by NaBPh4, followed by stannyl radical elimination.The reaction of a Lewis acid with a ruthenium complex, [Cp(CO)(ERィイD23ィエD2)Ru-{PN(Me)CHィイD22ィエD2CHィイD22ィエD2NMe(OMe)}] (where E = C, Si , Ge, Sn), which has both diamino-substituted phosphite and a group 14 element ligand was examined. Irrespective of the kind of E, an OMe group on the phosphorus is abstracted as an anion to give a phosphenium complex. The following reaction depends on the kind of E. If ERィイD23ィエD2 is an alkyl group, the ligand migrates from Ru to the phospheni um phosphorus, whereas if ERィイD23ィエD2 is a silyl or a germyl group, no migration reaction takes place. If ERィイD23ィエD2 is a stannyl group, an al kyl group on the tin migrates to the phosphenium phosphorus to give a stannylene complex, which has a double bond between Ru and Sn.The reaction of a Lewis acid with [Cp(CO)(SiMeィイD22ィエD2OMe)Fe{PN(Me)CHィイD22ィエD2CHィイD22ィエD2-NMe(OMe)}] containing an OR group on both coordinating P and Si atoms revealed that selective OR abstraction from Si occurs to give a silylene complex.

[Cp(CO)D22D2(SnD1nD1BuD23D2)W{PN(Me)CHD22D2CHD22D2NMe(OMe )}]与路易斯酸的反应揭示了磷鎓络合物的形成，随后D1BU基团从Sn迁移到P，得到如果该反应在 NaBPh D24D2 存在下发生，则会形成磷化物 Cp(CO)D22D2W{PN(Me)CHD22D2CHD22 D2NMe}，这种前所未有的反应被认为是 W-P 键级增加一。消除甲锡烷基和甲氧基。 NaBPh4 对鏻络合物进行电子还原，然后消除甲锡烷基。路易斯酸与钌络合物的反应，[Cp(CO)(ER-D23-D2)Ru-{PN(Me)CH- D22D2CHD22D2NMe( OMe)}]（其中 E = C、Si、Ge、Sn），其中都有二氨基取代检测亚磷酸酯和第 14 族元素配体，无论 E 的种类如何，磷上的 OMe 基团都会被提取为阴离子，得到鏻络合物。 D2 是烷基，配体从 Ru 迁移到磷鎓磷，而如果 D23 是如果 ERIID23IeD2 是甲锡基，则锡上的烷基会迁移到磷鎓上，形成亚锡络合物，该络合物在 Ru 和 Sn 之间具有双键。路易斯酸与在配位的 P 和 Si 原子上均含有 OR 基团的 [Cp(CO)(SiMeD22D2OMe)Fe{PN(Me)CHD22D2CHD22D22D2-NMe(OMe)}] 表明，Si 发生选择性 OR 夺取，得到亚硅基络合物。